Indictive currents on relay contact

Interesting. When I was a kid we did similar things with carbon microphones from old phones. Of course there I don't think it was a quantum tunneling effect, just that it conducted better when you squeezed everything together.

Tim.

None. AC is self-quenching.

John

You could use something like a MOC3030, opto-triac, with a zero-cross detector built in, to get v=0 turn-on.

Just trying the snubber across the relay contacts could help.

Cheers! Rich

Suck it and see ? 0.1uF and 100R in series is a classic off the shelf combination.

Triac.

Graham

A useless suggestion.... quantum tunneling capsules from peratech.

This is a particularly badly designed site so

some of the stuff is available from Maplin

martin

A 1N4004 diode across the inductive motor, so that, when the relay contact is closed, the diode is reversed biased, and when the contact is open, the inertia of the inductive current still finds a path to flow, through the diode. The diode should be as close as possible to the relay (not to the motor). With this, you can completely eliminate the spark.

MOSFETs.

Sorry, now I read your supply is AC.

Best is to turn on and off when i=0. A TRIAC solves for you half of the problem (it always turns off when i=0). If you want to also turn on when i=0, you need to detect it yourself. Even if you do all that, you still may need a snubber to avoid unwanted triggering of the TRIAC (due to excessive dV/dt, when the switch opens). A series RC network will do the job.

Beware of voltage leakage through snubber circuit. You may want to consider snubberless triacs like ST T820W. They have high dv/dt ratings

One day Joe G (Home) got dressed and committed to text

Is the arcing causing a problem ??

-- Regards ..... Rheilly Phoull

use a solid state relay, most of them turn off at or near the base line (zero crossing), but that time the back flow should be reduced to a min.. you could also use some TVS diodes.

Oh,

Supply is 110V AC

(I see why others complain about top-posting now)

yes, exactly like that.

Beware, however, the type of resistor you choose. If contact is made at the peak of AC line voltage (156V) and the capacitor is discharged, the

Ppeak = (155V^2)/100R = 243W

which will exponentially decay to zero with a time constant of 10us, for the suggested 0.1uF + 100R snubber. If you picked 10R, the peak power would be 10 times higher, but the time constant would be ten times shorter.

Make sure the resistor has a suitable voltage rating, and can cope with the peak pulse power (whatever that works out to be).

Also, the start-up current of the AC motor may be quite a bit higher than its operating current.